MAT137 | Calculus! | Lecture 8
Today:
Squeeze Theorem (v. 2.11-2.12)Trigonometric Limit (v. 2.17)
Next:
Watch videos on:Continuity (v. 2.13 - 2.16)
official website http://uoft.me/MAT137Beatriz Navarro-Lameda L0601 MAT137 03 October 2017
Bad theorem
Bad Theorem
Let f and g be functions with domain R, except perhaps a.IF lim
x→af (x) = 0,
THEN limx→a
f (x)g(x) = 0.
Bad Proof
limx→a
f (x)g(x) =[
limx→a
f (x)]·[
limx→a
g(x)]
= 0 ·[
limx→a
g(x)]
= 0, because 0
times anything is 0.
Find the error in the proof.
Show the theorem is false with a counterexample.
Beatriz Navarro-Lameda L0601 MAT137 03 October 2017
Limit: Bounded × something that goes to 0
Theorem
Let a ∈ R. Let f and g be functions with domain R, except perhaps a.
IF
limx→a
f (x) = 0
g is bounded.
THEN limx→a
f (x)g(x) = 0.
Definition (Bounded function)
Let g : R→ R. We say that g is bounded if
∃M > 0 s.t. ∀x ∈ R |g(x)| ≤ M.
Beatriz Navarro-Lameda L0601 MAT137 03 October 2017
The Squeeze Theorem
Theorem
Let p > 0. Suppose that for all x with 0 < |x − a| < p,
f (x) ≤ g(x) ≤ h(x).
IFlimx→a
f (x) = L and limx→a
h(x) = L,
THENlimx→a
g(x) = L.
Beatriz Navarro-Lameda L0601 MAT137 03 October 2017
The Squeeze Theorem
Exercise
Use the Squeeze Theorem to prove that:IF f (x) is bounded and lim
x→ag(x) = 0, THEN lim
x→af (x)g(x) = 0.
Exercise
IF limx→a|f (x)| = 0, THEN lim
x→af (x) = 0.
Hint: First show that −|f (x)| ≤ f (x) ≤ |f (x)|.
Beatriz Navarro-Lameda L0601 MAT137 03 October 2017
The Squeeze Theorem
Exercise
Use the Squeeze Theorem to prove that:IF f (x) is bounded and lim
x→ag(x) = 0, THEN lim
x→af (x)g(x) = 0.
Exercise
IF limx→a|f (x)| = 0, THEN lim
x→af (x) = 0.
Hint: First show that −|f (x)| ≤ f (x) ≤ |f (x)|.
Beatriz Navarro-Lameda L0601 MAT137 03 October 2017
Limits involving sin(1/x) Part I
The reason that limx→0
sin(1/x) does not exist is:
1 because no matter how close x gets to 0, there are x ’s near 0 forwhich sin(1/x) = 1, and some for which sin(1/x) = −1
2 because the function values oscillate around 0
3 because 1/0 is undefined
4 all of the above
-1.5 -1.0 -0.5 0.5 1.0 1.5
-1.0
-0.5
0.5
1.0
Beatriz Navarro-Lameda L0601 MAT137 03 October 2017
Limits involving sin(1/x) Part I
The reason that limx→0
sin(1/x) does not exist is:
1 because no matter how close x gets to 0, there are x ’s near 0 forwhich sin(1/x) = 1, and some for which sin(1/x) = −1
2 because the function values oscillate around 0
3 because 1/0 is undefined
4 all of the above
-1.5 -1.0 -0.5 0.5 1.0 1.5
-1.0
-0.5
0.5
1.0
Beatriz Navarro-Lameda L0601 MAT137 03 October 2017
Limits involving sin(1/x) Part II
The limit limx→0
x2 sin(1/x)
1 does not exist because no matter how close x gets to 0, there are x ’snear 0 for which sin(1/x) = 1, and some for which sin(1/x) = −1
2 does not exist because the function values oscillate around 0
3 does not exist because 1/0 is undefined
4 equals 0
5 equals 1
Beatriz Navarro-Lameda L0601 MAT137 03 October 2017
Limits involving sin(1/x) Part II
The limit limx→0
x2 sin(1/x)
1 does not exist because no matter how close x gets to 0, there are x ’snear 0 for which sin(1/x) = 1, and some for which sin(1/x) = −1
2 does not exist because the function values oscillate around 0
3 does not exist because 1/0 is undefined
4 equals 0
5 equals 1
Beatriz Navarro-Lameda L0601 MAT137 03 October 2017
Some Trigonometric Limits
Trig. Limit 1
You already know that limx→0
sin(x)
x= 1.
Use to evaluate limx→0
sin(3x)
x.
Answer:
1 1
2 3
3 13
-1.5 -1.0 -0.5 0.5 1.0 1.5
1
2
3
Beatriz Navarro-Lameda L0601 MAT137 03 October 2017
Some Trigonometric Limits
Trig. Limit 1
You already know that limx→0
sin(x)
x= 1.
Use to evaluate limx→0
sin(3x)
x.
Answer:
1 1
2 3
3 13
-1.5 -1.0 -0.5 0.5 1.0 1.5
1
2
3
Beatriz Navarro-Lameda L0601 MAT137 03 October 2017
Some Trigonometric Limits
Trig. Limit 2
The limit limx→0
1− cos(x)
xis
Answer:
1 1
2 0
3 12
Hint
Use the trigonometric identity sin2 x + cos2 x = 1.
Beatriz Navarro-Lameda L0601 MAT137 03 October 2017